Skip to main content
Mathematics LibreTexts

3.6: Solve Equations Using Integers; The Division Property of Equality

  • Page ID
    114886
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    Learning Objectives

    By the end of this section, you will be able to:

    • Determine whether an integer is a solution of an equation
    • Solve equations with integers using the Addition and Subtraction Properties of Equality
    • Model the Division Property of Equality
    • Solve equations using the Division Property of Equality
    • Translate to an equation and solve

    Be Prepared 3.10

    Before you get started, take this readiness quiz.

    Evaluatex+4whenx=−4.Evaluatex+4whenx=−4.
    If you missed this problem, review Example 3.22.

    Be Prepared 3.11

    Solve:y6=10.Solve:y6=10.
    If you missed this problem, review Example 2.33.

    Be Prepared 3.12

    Translate into an algebraic expression 55 less than x.x.
    If you missed this problem, review Table 1.3.

    Determine Whether a Number is a Solution of an Equation

    In Solve Equations with the Subtraction and Addition Properties of Equality, we saw that a solution of an equation is a value of a variable that makes a true statement when substituted into that equation. In that section, we found solutions that were whole numbers. Now that we’ve worked with integers, we’ll find integer solutions to equations.

    The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number or an integer.

    How To

    How to determine whether a number is a solution to an equation.

    1. Step 1. Substitute the number for the variable in the equation.
    2. Step 2. Simplify the expressions on both sides of the equation.
    3. Step 3. Determine whether the resulting equation is true.
      • If it is true, the number is a solution.
      • If it is not true, the number is not a solution.

    Example 3.60

    Determine whether each of the following is a solution of 2x5=−13:2x5=−13:

    1. x=4x=4
    2. x=−4x=−4
    3. x=−9.x=−9.
    Answer

    Substitute 4 for x in the equation to determine if it is true.
    .
    . .
    Multiply. .
    Subtract. .

    Since x=4x=4 does not result in a true equation, 44 is not a solution to the equation.

    Substitute −4 for x in the equation to determine if it is true. .
    . .
    Multiply. .
    Subtract. .

    Since x=−4x=−4 results in a true equation, −4−4 is a solution to the equation.

    Substitute −9 for x in the equation to determine if it is true.
    .
    Substitute −9 for x. .
    Multiply. .
    Subtract. .

    Since x=−9x=−9 does not result in a true equation, −9−9 is not a solution to the equation.

    Try It 3.119

    Determine whether each of the following is a solution of 2x8=−14:2x8=−14:

    1. x=−11x=−11
    2. x=11x=11
    3. x=−3x=−3

    Try It 3.120

    Determine whether each of the following is a solution of 2y+3=−11:2y+3=−11:

    1. y=4y=4
    2. y=−4y=−4
    3. y=−7y=−7

    Solve Equations with Integers Using the Addition and Subtraction Properties of Equality

    In Solve Equations with the Subtraction and Addition Properties of Equality, we solved equations similar to the two shown here using the Subtraction and Addition Properties of Equality. Now we can use them again with integers.

    This figure has two columns. The first column has the equation x plus 4 equals 12. Underneath there is x plus 4 minus 4 equals 12 minus 4. Under this there is x equals 8. The second column has the equation y minus 5 equals 9. Underneath there is the equation y minus 5 plus 5 equals 9 plus 5. Under this there is y equals 14.

    When you add or subtract the same quantity from both sides of an equation, you still have equality.

    Properties of Equalities

    Subtraction Property of Equality Addition Property of Equality
    For any numbersa,b,c,For any numbersa,b,c,
    ifa=bthenac=bc.ifa=bthenac=bc.
    For any numbersa,b,c,For any numbersa,b,c,
    ifa=bthena+c=b+c.ifa=bthena+c=b+c.

    Example 3.61

    Solve: y+9=5.y+9=5.

    Answer

    .
    Subtract 9 from each side to undo the addition. .
    Simplify. .

    Check the result by substituting −4−4 into the original equation.

    y+9=5y+9=5
    Substitute −4 for y −4+9=?5−4+9=?5
    5=55=5

    Since y=−4y=−4 makes y+9=5y+9=5 a true statement, we found the solution to this equation.

    Try It 3.121

    Solve:

    y+11=7y+11=7

    Try It 3.122

    Solve:

    y+15=−4y+15=−4

    Example 3.62

    Solve: a6=−8a6=−8

    Answer

    .
    Add 6 to each side to undo the subtraction. .
    Simplify. .
    Check the result by substituting −2−2 into the original equation: .
    Substitute −2−2 for aa .
    .

    The solution to a6=−8a6=−8 is −2.−2.

    Since a=−2a=−2 makes a6=−8a6=−8 a true statement, we found the solution to this equation.

    Try It 3.123

    Solve:

    a2=−8a2=−8

    Try It 3.124

    Solve:

    n4=−8n4=−8

    Model the Division Property of Equality

    All of the equations we have solved so far have been of the form x+a=bx+a=b or xa=b.xa=b. We were able to isolate the variable by adding or subtracting the constant term. Now we’ll see how to solve equations that involve division.

    We will model an equation with envelopes and counters in Figure 3.21.

    This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed.
    Figure 3.21

    Here, there are two identical envelopes that contain the same number of counters. Remember, the left side of the workspace must equal the right side, but the counters on the left side are “hidden” in the envelopes. So how many counters are in each envelope?

    To determine the number, separate the counters on the right side into 22 groups of the same size. So 66 counters divided into 22 groups means there must be 33 counters in each group (since 6÷2=3).6÷2=3).

    What equation models the situation shown in Figure 3.22? There are two envelopes, and each contains xx counters. Together, the two envelopes must contain a total of 66 counters. So the equation that models the situation is 2x=6.2x=6.

    This image has two columns. In the first column are two identical envelopes. In the second column there are six blue circles, randomly placed. Under the figure is two times x equals 6.
    Figure 3.22

    We can divide both sides of the equation by 22 as we did with the envelopes and counters.

    This figure has two rows. The first row has the equation 2x divided by 2 equals 6 divided by 2. The second row has the equation x equals 3.

    We found that each envelope contains 3 counters.3 counters. Does this check? We know 2·3=6,2·3=6, so it works. Three counters in each of two envelopes does equal six.

    Figure 3.23 shows another example.

    This image has two columns. In the first column are three envelopes. In the second column there are four rows of  three blue circles. Underneath the image is the equation 3x equals 12.
    Figure 3.23

    Now we have 3Figure 3.24.

    This image has two columns. In the first column are four envelopes. In the second column there are twelve blue circles.
    Figure 3.24

    The equation that models the situation is 3x=12.3x=12. We can divide both sides of the equation by 3.3.

    This image shows the equation 3x divided by 3 equals 12 divided by 3. Below this equation is the equation x equals 4.

    Does this check? It does because 3·4=12.3·4=12.

    Manipulative Mathematics

    Example 3.63

    Write an equation modeled by the envelopes and counters, and then solve it.

    This image has two columns. In the first column are four envelopes. In the second column there are 8 blue circles.
    Answer

    There are 4 envelopes,4 envelopes, or 44 unknown values, on the left that match the 8 counters8 counters on the right. Let’s call the unknown quantity in the envelopes x.x.

    Write the equation. .
    Divide both sides by 4. .
    Simplify. .

    There are 2 counters2 counters in each envelope.

    Try It 3.125

    Write the equation modeled by the envelopes and counters. Then solve it.

    This image has two columns. In the first column are four envelopes. In the second column there are 12 blue circles.

    Try It 3.126

    Write the equation modeled by the envelopes and counters. Then solve it.

    This image has two columns. In the first column are three envelopes. In the second column there are six blue circles.

    Solve Equations Using the Division Property of Equality

    The previous examples lead to the Division Property of Equality. When you divide both sides of an equation by any nonzero number, you still have equality.

    Division Property of Equality

    For any numbers a,b,c,and c0, If a=bthen ac=bc. For any numbers a,b,c,and c0, If a=bthen ac=bc.

    Example 3.64

    Solve:7x=−49.Solve:7x=−49.

    Answer

    To isolate x,x, we need to undo multiplication.

    .
    Divide each side by 7. .
    Simplify. .

    Check the solution.

    7x=−497x=−49
    Substitute −7 for x. 7(−7)=?−497(−7)=?−49
    −49=−49−49=−49

    Therefore, −7−7 is the solution to the equation.

    Try It 3.127

    Solve:

    8a=568a=56

    Try It 3.128

    Solve:

    11n=12111n=121

    Example 3.65

    Solve: −3y=63.−3y=63.

    Answer

    To isolate y,y, we need to undo the multiplication.

    .
    Divide each side by −3. .
    Simplify .

    Check the solution.

    −3y=63−3y=63
    Substitute −21 for y. −3(−21)=?63−3(−21)=?63
    63=6363=63

    Since this is a true statement, y=−21y=−21 is the solution to the equation.

    Try It 3.129

    Solve:

    −8p=96−8p=96

    Try It 3.130

    Solve:

    −12m=108−12m=108

    Translate to an Equation and Solve

    In the past several examples, we were given an equation containing a variable. In the next few examples, we’ll have to first translate word sentences into equations with variables and then we will solve the equations.

    Example 3.66

    Translate and solve: five more than xx is equal to −3.−3.

    Answer

    five more than xx is equal to −3−3
    Translate x+5=−3x+5=−3
    Subtract 55 from both sides. x+55=−35x+55=−35
    Simplify. x=−8x=−8

    Check the answer by substituting it into the original equation.

    x + 5 = −3 −8 + 5 = ? −3 −3 = −3 x + 5 = −3 −8 + 5 = ? −3 −3 = −3

    Try It 3.131

    Translate and solve:

    Seven more than xx is equal to −2−2.

    Try It 3.132

    Translate and solve:

    Eleven more thanyis equal to 2.Eleven more thanyis equal to 2.

    Example 3.67

    Translate and solve: the difference of nn and 66 is −10.−10.

    Answer

    the difference of nn and 66 is −10−10
    Translate. n6=−10n6=−10
    Add 66 to each side. n6+6=−10+6n6+6=−10+6
    Simplify. n=−4n=−4

    Check the answer by substituting it into the original equation.

    n 6 = −10 −4 6 = ? −10 −10 = −10 n 6 = −10 −4 6 = ? −10 −10 = −10

    Try It 3.133

    Translate and solve:

    The difference of pp and 22 is −4−4.

    Try It 3.134

    Translate and solve:

    The difference of qq and 77 is −3−3.

    Example 3.68

    Translate and solve: the number 108108 is the product of −9−9 and y.y.

    Answer

    the number of 108108 is the product of −9−9 and yy
    Translate. 108=−9y108=−9y
    Divide by −9−9. 108−9=−9y−9108−9=−9y−9
    Simplify. −12=y−12=y

    Check the answer by substituting it into the original equation.

    108 = −9 y 108 = ? −9 ( −12 ) 108 = 108 108 = −9 y 108 = ? −9 ( −12 ) 108 = 108

    Try It 3.135

    Translate and solve:

    The number 132132 is the product of −12−12 and yy.

    Try It 3.136

    Translate and solve:

    The number 117117 is the product of −13−13 and zz.

    Media

    Section 3.5 Exercises

    Practice Makes Perfect

    Determine Whether a Number is a Solution of an Equation

    In the following exercises, determine whether each number is a solution of the given equation.

    285.

    4x2=64x2=6

    1. x=−2x=−2
    2. x=−1x=−1
    3. x=2x=2
    286.

    4y10=−144y10=−14

    1. y=−6y=−6
    2. y=−1y=−1
    3. y=1y=1
    287.

    9a+27=−639a+27=−63

    1. a=6a=6
    2. a=−6a=−6
    3. a=−10a=−10
    288.

    7c+42=−567c+42=−56

    1. c=2c=2
    2. c=−2c=−2
    3. c=−14c=−14

    Solve Equations Using the Addition and Subtraction Properties of Equality

    In the following exercises, solve for the unknown.

    289.

    n + 12 = 5 n + 12 = 5

    290.

    m + 16 = 2 m + 16 = 2

    291.

    p + 9 = −8 p + 9 = −8

    292.

    q + 5 = −6 q + 5 = −6

    293.

    u 3 = −7 u 3 = −7

    294.

    v 7 = −8 v 7 = −8

    295.

    h 10 = −4 h 10 = −4

    296.

    k 9 = −5 k 9 = −5

    297.

    x + ( −2 ) = −18 x + ( −2 ) = −18

    298.

    y + ( −3 ) = −10 y + ( −3 ) = −10

    299.

    r ( −5 ) = −9 r ( −5 ) = −9

    300.

    s ( −2 ) = −11 s ( −2 ) = −11

    Model the Division Property of Equality

    In the following exercises, write the equation modeled by the envelopes and counters and then solve it.

    301.
    No Alt Text
    302.
    No Alt Text
    303.
    No Alt Text
    304.
    No Alt Text

    Solve Equations Using the Division Property of Equality

    In the following exercises, solve each equation using the division property of equality and check the solution.

    305.

    5 x = 45 5 x = 45

    306.

    4 p = 64 4 p = 64

    307.

    −7 c = 56 −7 c = 56

    308.

    −9 x = 54 −9 x = 54

    309.

    −14 p = −42 −14 p = −42

    310.

    −8 m = −40 −8 m = −40

    311.

    −120 = 10 q −120 = 10 q

    312.

    −75 = 15 y −75 = 15 y

    313.

    24 x = 480 24 x = 480

    314.

    18 n = 540 18 n = 540

    315.

    −3 z = 0 −3 z = 0

    316.

    4 u = 0 4 u = 0

    Translate to an Equation and Solve

    In the following exercises, translate and solve.

    317.

    Four more than nn is equal to 1.

    318.

    Nine more than mm is equal to 5.

    319.

    The sum of eight and pp is −3−3.

    320.

    The sum of two and qq is −7−7.

    321.

    The difference of aa and three is −14−14.

    322.

    The difference of bb and 55 is −2−2.

    323.

    The number −42 is the product of −7 and xx.

    324.

    The number −54 is the product of −9 and yy.

    325.

    The product of -15 and ff is 75.

    326.

    The product of −18 and gg is 36.

    327.

    −6 plus cc is equal to 4.

    328.

    −2 plus dd is equal to 1.

    329.

    Nine less than mm is −4.

    330.

    Thirteen less than nn is −10−10.

    Mixed Practice

    In the following exercises, solve.

    331.
    1. x+2=10x+2=10
    2. 2x=102x=10
    332.
    1. y+6=12y+6=12
    2. 6y=126y=12
    333.
    1. −3p=27−3p=27
    2. p3=27p3=27
    334.
    1. −2q=34−2q=34
    2. q2=34q2=34
    335.

    a 4 = 16 a 4 = 16

    336.

    b 1 = 11 b 1 = 11

    337.

    −8 m = −56 −8 m = −56

    338.

    −6 n = −48 −6 n = −48

    339.

    −39 = u + 13 −39 = u + 13

    340.

    −100 = v + 25 −100 = v + 25

    341.

    11 r = −99 11 r = −99

    342.

    15 s = −300 15 s = −300

    343.

    100 = 20 d 100 = 20 d

    344.

    250 = 25 n 250 = 25 n

    345.

    −49 = x 7 −49 = x 7

    346.

    64 = y 4 64 = y 4

    Everyday Math

    347.

    Cookie packaging A package of 51 cookies51 cookies has 33 equal rows of cookies. Find the number of cookies in each row, c,c, by solving the equation 3c=51.3c=51.

    348.

    Kindergarten class Connie’s kindergarten class has 24 children.24 children. She wants them to get into 44 equal groups. Find the number of children in each group, g,g, by solving the equation 4g=24.4g=24.

    Writing Exercises

    349.

    Is modeling the Division Property of Equality with envelopes and counters helpful to understanding how to solve the equation 3x=15?3x=15? Explain why or why not.

    350.

    Suppose you are using envelopes and counters to model solving the equations x+4=12x+4=12 and 4x=12.4x=12. Explain how you would solve each equation.

    351.

    Frida started to solve the equation −3x=36−3x=36 by adding 33 to both sides. Explain why Frida’s method will not solve the equation.

    352.

    Raoul started to solve the equation 4y=404y=40 by subtracting 44 from both sides. Explain why Raoul’s method will not solve the equation.

    Self Check

    After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

    .

    Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?


    This page titled 3.6: Solve Equations Using Integers; The Division Property of Equality is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax.

    • Was this article helpful?